rms values of rectifier waveforms

8/18/2019 RMS values of rectifier waveforms

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-./0&12.,1 +3 %4.(1&'(045 6+2/71.&5 0,8 %,.&9: %,9',..&',

;,'


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Fundamentals of Power Electronics 77 Chapter 18: PWM Rectifiers

!"#V S6; 904(*8 .@ )*3,


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!"#$ &'()*+,- *'..). /,( )00+1+),12+, 33& 4+-4567/*+82 9)18+0+)9.

Objective: extend procedure of Chapter 3, to predict the output

voltage, duty cycle variations, and efficiency, of PWM CCM low

harmonic rectifiers.

Approach: Use the models developed in Chapter 3. Integrate over

one ac line cycle to determine steady-state waveforms and average

power.

Boost example

+ –

Q1

L

C R

+

v(t)

–

D1

vg(t)

ig(t) R

L i(t)

+ –

R

+

v(t)

–

vg(t)

ig(t) R L

i(t) DRon

+ –

D' : 1V F

Dc-dc boost converter circuit Averaged dc model


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&'()*+,- 84) /15(1 :''.8 9)18+0+)9

Rvac(t)

iac(t)+

vg(t)

–

ig(t)

+

v(t)

–

id (t)

Q1

L

C

D1

controller

i(t)

R L

+ –

R

+

v(t) = V

–

vg(t)

ig(t) R L

i(t) = I d(t) Ron

+ –

d'(t) : 1V F

id (t)

C (large)

Boost

rectifier

circuit

Averaged

model


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?@/>A*)B :''.8 9)18+0+)9


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!"#$#! ?@A9)..+', 0'9 1',89'**)9 (782 121*) !"#$

+ –

R

+

v(t) = V

–

vg(t)

ig(t) i(t) = I

d(t) Ron

d'(t) : 1

id (t)

C (large)

Solve input side of

model:

ig(t )d (t ) Ron = vg(t ) – d '(t )v

with ig(t ) =vg(t )

Re

eliminate ig(t ):

vg t

Red (t ) Ron = vg(t ) – d '(t )v

vg(t ) = V M sin !t

solve for d (t ):

d (t ) =v – vg(t )

v – vg(t ) Ron Re

Again, these expressions neglect converter dynamics, and assume

that the converter always operates in CCM.


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!"#$#G ?@A9)..+', 0'9 84) (1 *'/( 1799),8

+ –

R

+

v(t) = V

–

vg(t)

ig(t) i(t) = I

d(t) Ron

d'(t) : 1

id (t)

C (large)

Solve output side of

model, using chargebalance on capacitor C :

I = id T ac

id (t ) = d '(t )ig(t ) = d '(t )vg t

Re

Butd’(t ) is:

d '(t ) =

vg(t ) 1 – Ron Re

v – vg(t ) Ron Re

hence id (t ) can be expressed as

id (t ) =vg

2(t )

Re

1 – Ron Re

v – vg(t ) Ron Re

Next, average id

(t ) over an ac line period, to find the dc load current I .


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H1 *'/( 1799),8 %

I = id T ac= 2

T ac

V M 2

Re

1 – Ron Re

sin2 !t

v – V M Ron

Resin !t

dt

0

T ac/2

Now substitute vg (t ) = V M sin &t , and integrate to find "id (t )#T ac:

This can be written in the normalized form

I = 2T ac

V M 2

VRe1 –

Ron Re

sin2 !t

1 – a sin !t dt

0

T ac/2

with a = V M

V

Ron Re


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By waveform symmetry, we need only integrate from 0 to T ac/4. Also,make the substitution % = &t :

I = V M

2

VRe1 –

Ron Re

2*

sin2 -

1 – a sin -d -

0

*/2

This integral is obtained not only in the boost rectifier, but also in the

buck-boost and other rectifier topologies. The solution is

4*

sin2 -

1 – a sin -d -

0

*/2

= F (a) = 2a2*

– 2a – * +4 sin

– 1a + 2 cos – 1 a

1 – a 2

• Result is in closed form

• a is a measure of the loss

resistance relative to Re

• a is typically much smaller than

unity


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F4) +,8)-9/* &J'K

F(a)

a

–0.15 –0.10 –0.05 0.00 0.05 0.10 0.15

0.85

0.9

0.95

1

1.05

1.1

1.15

4*

sin2 -

1 – a sin -d -

0

*/2

= F (a) = 2a2*

– 2a – * +4 sin

– 1a + 2 cos – 1 a

1 – a 2

F (a) & 1 + 0.862a + 0.78a2

Approximation via

polynomial:

For | a | ' 0.15, thisapproximate expression iswithin 0.1% of the exact

value. If the a2 term is

omitted, then the accuracydrops to ±2% for | a | ' 0.15.The accuracy of F(a)

coincides with the accuracy

of the rectifier efficiency (.


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!"#$#L D'*78+', 0'9 1',=)98)9 )00+1+),12 (

Converter average input power is

Pin = pin(t ) T ac=

V M 2 Re

Average load power is

Pout = VI = V V M

2

VRe1 –

Ron Re

F (a)

2with a =

V M V

Ron Re

So the efficiency is

/ = Pout

Pin= 1 –

Ron Re

F (a)

Polynomial approximation:

/ & 1 – Ron Re

1 + 0.862 V M

V

Ron Re

+ 0.78 V M

V

Ron Re

2


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;''.8 9)18+0+)9 )00+1+),12

/ = Pout

Pin= 1 –

Ron Re

F (a)

0.0 0.2 0.4 0.6 0.8 1.0

0.75

0.8

0.85

0.9

0.95

1

V M /V

/

Ron/ Re

= 0. 05

R o n/ Re

= 0. 1

R o n / R e

= 0. 1 5

R o n / R e = 0

. 2

• To obtain highefficiency, choose V

slightly larger than V M

• Efficiencies in the range90% to 95% can then beobtained, even with Ronas high as 0.2 Re

• Losses other than

MOSFET on-resistance

are not included here


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Let us design for a given efficiency. Consider the following

specifications:Output voltage 390 V

Output power 500 W

rms input voltage 120 V

Efficiency 95%

Assume that losses other than the MOSFET conduction loss are

negligible.

Average input power is

Pin = out

/ = 500 W

0.95 = 526 W

Then the emulated resistance is

Re =V g, rms

2

Pin=

(120 V)2

526 W = 27.4 0


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H).+-, )@/>A*)

Also, M V

= 120 2 V390 V

= 0.435

0.0 0.2 0.4 0.6 0.8 1.0

0.75

0.8

0.85

0.9

0.95

1

V M /V

/

Ron/ Re

= 0. 05

R o n/ Re

= 0. 1

R o n / R e

= 0. 1 5

R o n / R e

= 0. 2

95% efficiency with

V M /V = 0.435 occurs

with Ron/ Re ) 0.075.

So we require a

MOSFET with on

resistance of

Ron ) (0.075) Re

= (0.075) (27.4 0) = 2 0

rms values of rectifier waveforms

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